In this case I would use the method of synthetic substitution, but since I never learned this technique 'til this year in Precalculus, let's stick with long division:

1) Write out the problem like an elementary division problem, with the bar over the 4m^2-9m^2...+7 and the m-3 to the side.

2) For the beginning, calculate how many times m (not m-3, just m) goes into 4m^3. In simpler terms, what would you multiply m by to get 4m^3? You would get 4m^2, so write that on the bar above.

3) Like regular division, write the product of m-3 and 4m^2 underneath the problem, drop another set of the equation down (just ONE expression, which, in this case would be the -9m^2 next). So far, you should have 4m^2 + 12m^2 written underneath to subtract from the first two terms in the dividend.

4) Repeat the process until you can go no further, which is when you reach a number without any variables. This (divided by m-3) will be your remainder.

I hope this makes sense....it'd be easier to show if I could show in person. Hope this helps! :)